Question: Simplify the following expression: $\dfrac{14y^5}{21y}$ You can assume $y \neq 0$.
$ \dfrac{14y^5}{21y} = \dfrac{14}{21} \cdot \dfrac{y^5}{y} $ To simplify $\frac{14}{21}$ , find the greatest common factor (GCD) of $14$ and $21$ $14 = 2 \cdot 7$ $21 = 3 \cdot 7$ $ \mbox{GCD}(14, 21) = 7 $ $ \dfrac{14}{21} \cdot \dfrac{y^5}{y} = \dfrac{7 \cdot 2}{7 \cdot 3} \cdot \dfrac{y^5}{y} $ $\phantom{ \dfrac{14}{21} \cdot \dfrac{5}{1}} = \dfrac{2}{3} \cdot \dfrac{y^5}{y} $ $ \dfrac{y^5}{y} = \dfrac{y \cdot y \cdot y \cdot y \cdot y}{y} = y^4 $ $ \dfrac{2}{3} \cdot y^4 = \dfrac{2y^4}{3} $